Question

If an iron ball and a wooden ball of the same radius are released from a height h in vacuum, then time taken by both of them to reach ground will be: 

1. Unequal
2. Zero
3. Exactly equal
4. Roughly equal

Answer 1

Correct Answer - Option 3 : Exactly equal

CONCEPT:

Equation of Kinematics:

• These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

⇒ V = U + at

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)

⇒ V2 =U2+ 2as

where, U = Initial velocity, V = Final velocity, g = Acceleration due to gravity, t = time, h = height/Distance covered and a = g

CALCULATION:

Given U_I = 0 m/s, U_W = 0 m/s, s_I = h, s_W = h and g = gravitational acceleration

Where UI = initial velocity of the iron ball, UW = initial velocity of the wooden ball, sI = initial height of the iron ball and sW = initial height of the wooden ball

• The second equation of motion is given as,

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)     -----(1)

• Since both the balls are released in a vacuum, so there will be no force other than gravity will act on the iron ball and wooden ball. So the acceleration of both the balls will be equal to the gravitational acceleration.
• By equation 1 it is clear that the displacement of a uniformly accelerated object does not depend on its mass, shape, and size.
• Let t_I and t_W be the time required by the iron and the wooden ball to reach the ground.

By equation 1 for the iron ball,

\(⇒ s_{I} =U_{I}t_{I}+\frac{1}{2}{gt_{I}^{2}}\)

\(⇒ h =(0\times t_{I})+\frac{1}{2}{gt_{I}^{2}}\)

\(⇒ t_{I}=\sqrt{\frac{2h}{g}}\)     -----(2)

By equation 1 for the wooden ball,

\(⇒ s_{W} =U_{W}t_{W}+\frac{1}{2}{gt_{W}^{2}}\)

\(⇒ h =(0\times t_{W})+\frac{1}{2}{gt_{W}^{2}}\)

\(⇒ t_{W}=\sqrt{\frac{2h}{g}}\)     -----(3)

By equation 2 and equation 3,

⇒ t_I = t_W

• Both the balls will reach the ground in equal time.
• Hence, option 3 is correct.